Homework #4 Solutions Due: July 3, Do the following exercises from Lax: Page 124: 9.1, 9.3, PDF Free Download

Do the following exercises from Lax: Page 124: 9.1, 9.3, 9.5 9.1. a) Find the number of different squares with vertices colored red, white, or blue. b) Find the number of different m-colored squares for any m. Solution. The symmetry group of a square is D 4 = { ε, α, α 2, α 3, β, αβ, α 2 β, α 3 β } and the set X on which we want D 4 = G to act is the set of all functions from the vertices of the square to the set of m colors. Identifying the elements of D 4 with permutations of the vertices of the square, as described on page 114, we can compute the cycle type of each permutation determined by g D 4 and then compute X g using Theorem 5.5.5, page 231 of the supplement (or corollary 10.3, page 127 of Lax). The results are, for m-colors: (l(ρ) denotes the number of cycles in the disjoint cycle factorization of ρ) ρ Cycle Rep. l(ρ) X ρ ε (1)(2)(3)(4) 4 m 4 α (1 2 3 4) 1 m α 2 (1 3)(2 4) 2 m 2 α 3 (1 4 3 2) 1 m β (1 2)(3 4) 2 m 2 αβ (1 3)(2)(4) 3 m 3 α 2 β (1, 4)(2, 3) 2 m 2 α 3 β (1)(2 4)(3) 3 m 3 Hence, Burnside s Lemma give the number of orbits as N = 1 8 (m4 + 2m 3 + 3m 2 + 2m). This is the answer to part b). To get part a), let m = 3 to get N = 1 168 (81 + 54 + 27 + 6) = 8 8 = 21. 9.3. Find the number of different regular hexagons with vertices colored red or blue. Solution. This is a coloring problem with m = 2 colors. The symmetry group is the symmetry group of a regular hexagon, that is, the dihedral group D 6 of degree 6, which consists of 6 rotations and 6 reflections. If the vertices are labeled from 1 to 6 counterclockwise, then the elements of D 6 can be represented by permutations in S 6. Math 4023 1

The set X on which G = D 6 acts is the set of all functions from the set of vertices of the hexagon to the set {red, blue}. The table of sizes of fixed point sets is ρ l(ρ) X ρ (1)(2)(3)(4)(5)(6) 6 2 6 (1 2 3 4 5 6) 1 2 1 (1 3 5)(2 4 6) 2 2 2 (1 4)(2 5)(3 6) 3 2 3 (1 5 3)(2 6 4) 2 2 2 (1 6 5 4 3 2) 1 2 1 (1 6)(2 5)(3 4) 3 2 3 (1 2)(3 6)(4 5) 3 2 3 (1 4)(2 3)(5 6) 3 2 3 (1)(4)(2 6)(3 5) 4 2 4 (2)(5)(1 3)(4 6) 4 2 4 (3)(6)(1 5)(2 4) 4 2 4 The number of different color patterns is the number of orbits of G acting on X, which by Burnside s Theorem is N = (2 6 + 3 2 4 + 4 2 3 + 2 2 2 + 2 2)/12 = 13. 9.5. A wheel is divided evenly into six different compartments. Each compartment can be painted red or white. The back of the wheel is black. How many different color wheels are there? Solution. This is similar to the previous problem except that instead of vertices, you are asked to color six equal wedges of the circle, that we shall label 1 to 6 in a counterclockwise direction. Since the back of the wheel is black, it follows that the only symmetries are rotations. A reflection is equivalent to turning the wheel over across a diagonal, but this would turn the back to the front and the back is different. Thus, the symmetry group is the cyclic group G = C 6 = (α) of rotations by multiples of 60 degrees, so that G = 6. This corresponds to the first six rows of the table in problem 9.3, when viewed as permutations. Hence, the number of distinct patterns is, according to Burnside s Lemma: N = 1 6 (26 + 2 3 + 2 2 2 + 2 2) = 14. 1. (a) Suppose that G is a group of order 96 and that H is a subgroup of G such that there are 6 left cosets of H in G. What is H? Solution. Lagrange s Theorem states that G = [G : H] H where the index [G : H] is the number of left cosets of H in G. Thus, H = G /[G : H] = 96/6 = 16. Math 4023 2

(b) Suppose that K and L are subgroups of a group M and assume that the following data are given: K = 9, L = 12, M < 100. What are the possible values of M? Solution. From Lagrange, it follows that M is a multiple of 9 and also a multiple of 12. Thus, M is a common multiple of 9 and 12. The least common multiple of 9 and 12 is 36, so the common multiples of 9 and 12 that are < 100 are 36 and 72, which are the possible values of M. 2. In this exercise G will denote the group Z 60 of multiplicatively invertible congruence classes of Z 60, with the group operation being multiplication of congruence classes modulo 60. (a) Verify that H = {[1] 60, [13] 60, [37] 60, [49] 60 } is a subgroup of G. Solution. The easiest way to verify that H is a subgroup is to write out the multiplication table (modulo 60): 1 13 37 49 1 1 13 37 49 13 13 49 1 37 37 37 1 49 13 49 49 37 13 1 From this table, it is clear the H is closed under multiplication modulo 60, and hence it is a subgroup. (b) How many distinct left cosets does H have in G? Solution. G = Z 60 = φ(60) = φ(4 3 5) = 2 2 4 = 16. By Lagrange s theorem, the number of left cosets of H in G is G / H = 16/4 = 4. 3. Let σ = ( 1 2 3 4 2 4 1 3 ) and τ = ( 1 2 3 4 3 4 1 2 ). In each case, solve the given group equation for χ S 4. (a) σχ = τ (b) χτ = σ (c) χτσ = id Solution. (a) χ = σ 1 τ = ( 1 2 3 4 3 1 4 2 ) ( 1 2 3 4 3 4 1 2 ) = ( 1 2 3 4 4 2 3 1 ). (b) χ = στ 1 = ( 1 2 3 4 2 4 1 3 ) ( 1 2 3 4 2 4 1 3 ) = ( 1 2 3 4 1 3 2 4 ). (c) χ = σ 1 τ 1 = σ 1 τ = ( 1 2 3 4 4 2 3 1 ). 4. Factor each of the following permutations into disjoint cycles. Using your answer express the inverse as a product of disjoint cycles. ( ) 1 2 3 4 5 6 7 8 9 (a) 4 7 9 8 2 1 6 3 5 Math 4023 3

(b) Solution. ( 1 4 8 3 9 5 2 7 6 ) ( 1 2 3 4 5 6 7 8 ) 9 6 4 8 9 3 1 7 5 2 Solution. ( 1 6 ) ( 2 4 9 ) ( 3 8 5 ) (c) ( 1 3 ) ( 2 5 7 ) ( 3 8 5 ) Solution. ( 1 3 8 7 2 5 ) (d) ( 3 5 7 ) ( 2 3 4 7 6 ) ( 3 5 7 ) 1 Solution. ( 2 5 4 3 6 ) 5. (a) How many permutations in S 5 fix 1? Solution. A permutation of {1, 2, 3, 4, 5} that fixes 1 is identified with a permutation of {2, 3, 4, 5}. The number of such permutations is the order of S 4, that is 4! = 24. (b) How many permutations in S 5 fix both 1 and 3? Solution. A permutation that fixes both 1 and 3 is identified with a permutation of the set {2, 4, 5}. There are 3! = 6 such permutations. 6. Let G = S 4 be the group of permutations of the set {1, 2, 3, 4}. For the purposes of this exercise it will probably be most convenient to write all elements of G in the disjoint cycle format. (a) Give a list of all of the distinct cyclic subgroups of G of order 4. Solution. A cyclic subgroup of order 4 is generated by an element of order 4. The order of a permutation is the least common multiple of the orders of the cycles in the disjoint cycle factorization of the permutation. The only way to get 4 as the least common multiple is to have a single cycle of length 4. Thus, the subgroups of order 4 of S 4 are those generated by a 4-cycle. Each such subgroup will have two generators, namely σ and σ 1. Note that there are only 6 distinct 4-cycles in S 4. This is because a 4-cycle in S 4 has the form (a 1 a 2 a 3 a 4 ) where a 1, a 2, a 3, a 4 is just a rearrangement of 1, 2, 3, 4. But any element in the 4-cycle can be taken as the first element listed so always take a 1 = 1. Thus, the number of ways of arranging the last 3 numbers is 3! = 6. Each subgroup contains exactly 2 of these 4-cycles so there are a total Math 4023 4

of 3 distinct cyclic subgroups of order 4. Hence, the distinct cyclic subgroups of order 4 in S 4 are: H 1 = {ε, (1 2 3 4), (1 3)(2 4), (1 4 3 2)} H 2 = {ε, (1 2 4 3), (1 4)(2 3), (1 3 4 2)} H 3 = {ε, (1 3 2 4), (1 2)(3 4), (1 4 2 3)}. (b) Given an example of a subgroup H of G of order 4 that is not cyclic. Solution. One example is to take a pair of disjoint 2-cycles and consider the group that they generate. For example, let H = {(1 2), (3 4), (1 2)(3 4)}. It is easy to check that H is closed under multiplication and hence is a subgroup of S 4. Moreover, the square of every element is the identity, so it is not cyclic (there is not element of order 4). 7. A rectangular design consists of 11 parallel stripes of equal width. If each stripe can be painted red, blue, or green, find the number of possible patterns. Solution. This shape is like the Polya s necktie problem in Example 5.5.3, Page 228. In our case, n = 11. The group of symmetries is the group G = {ε, α} where α is the rotation about the center of the rectangle by 180 degrees. Labeling the stripes from the left to the right as 1 to 11, the rotation α corresponds to the permutation α = (1 11)(2 10)(3 9)(4 8)(5 7)(6) so that l(α) = 6. The identity is of course a product of 11 1-cycles. Since there are 3 possible colors, we conclude that X is the set of all functions from the set of 11 stripes to the set of 3 colors. Hence, X ε = 3 11 and X α = 3 6. Thus, the number of distinct patterns is N = 1 2 (311 + 3 6 ) = 177876 = 88938. 2 8. Find the number of distinct bracelets consisting of six beads, where each bead is red, blue, or white. Solution. This is essentially a coloring problem for a regular hexagon with m = 3 colors. Thus it is essentially the same analysis as Problem 9.3. The symmetry group is the symmetry group of a regular hexagon, that is, the dihedral group D 6 of degree 6, which consists of 6 rotations and 6 reflections. If the vertices are labeled from 1 to 6 counterclockwise, then the elements of D 6 can be represented by permutations in S 6. Math 4023 5

Homework #4 Solutions Due: July 3, Do the following exercises from Lax: Page 124: 9.1, 9.3, 9.5